## Calculus (3rd Edition)

$18.93 \ m$
Let $x$ be the displacement of the particle. Therefore, $v(t)=\dfrac{dx}{dt}=0.02t^2+t$ Integrate the above equation to obtain the displacement of the particle over the time interval $[1,6]$: $\int_1^{6} dx=\int_1^{6} (0.02t^2+t) \ dt$ or, $=[\dfrac{0.02t^3}{3}-\dfrac{t^2}{2}]_1^{6}$ or, $=[\dfrac{0.02(6)^3}{3}-\dfrac{(6)^2}{2}]-[\dfrac{0.02(1)^3}{3}-\dfrac{(1)^2}{2}]$ or, $=19.44-0.806$ or, $=18.93 \ m$ Therefore, the displacement of the particle over the time interval $[1,6]$ is $18.93 \ m$